# Search results

1. ### Heat and Phase change: latent heat

I'm still not sure how to find the melting of the ice, do i need to use latent heat constant? ahhh so add the two values together and then multiply by the area*the density?
2. ### Heat and Phase change: latent heat

good after I find that I need add that to melting the ice. How does the area and the density come into play?
3. ### Heat and Phase change: latent heat

So I need to use the formula Q=mass*specific heat*change in temp for the first part right?
4. ### Heat and Phase change: latent heat

Specific heat for water=4186
5. ### Heat and Phase change: latent heat

is the latent fusion the same thing as the temperature for which ice needs to be in order to melt?
6. ### Heat and Phase change: latent heat

well the melting point is at 0. degrees celcius
7. ### Heat and Phase change: latent heat

1.if i raise the temp of the ice to the melting pt. it would need to be at its boiling pt 100 degrees celcius. heat rises 2. now must melt the ice at 0 degrees celcius I'm sorry I really don't know what I'm doing just learned this for the first time today along with four other physics chapter...
8. ### Heat and Phase change: latent heat

Homework Statement A woman finds the front windshield of her car covered with ice at -12.8°C. The ice has a thickness of 4.60 10-4 m, and the windshield has an area of 1.25 m2. The density of ice is 917 kg/m3. How much heat is required to melt the ice? Homework Equations Q=mL or change...
9. ### Temperature conversion

sometimes i just try to complicate problems...thanks for your help I appreciate it very much ;-)
10. ### Temperature conversion

o sorry, When you convert these to Fahrenheit degrees, it means 176.13 Fahrenheit degrees above freezing, where freezing is 32 F. so, Tf = (9/5)*Tc+32
11. ### Temperature conversion

Man I was making this problem much tougher than it was! thanks for your help!
12. ### Temperature conversion

first i would take the celcius and subtract 32 degrees since its above the farenheight freezing pt. and then convert to farenheight by multiplying by (9/5)....right....and in the other part i would add 32 degrees since its below the zero pt??
13. ### Temperature conversion

Oops So I need to subtract 32 degrees from each answer to take an account for the freezing pt.
14. ### Temperature conversion

Homework Statement On the moon the surface temperature ranges from 371 K during the day to 1.09 102 K at night. (a) What are these temperatures on the Celsius scale? (b) What are these temperatures on the Fahrenheit scale? Homework Equations T=Tc+273.15 T=Kelvin temperature Tc=Celcius...
15. ### Need a little help!

that is much easier to understand!
16. ### Need a little help!

Its equal to the angular velocity so would it be 118*(.050/.030)
17. ### Need a little help!

So I'm trying to learn how to do it using the angular velocity way. I'm not familiar with this method and I was hoping if you could check my work. a = omega^2 r So a is proportional to r. a(r) = a(@ r') (r / r') plug the numbers in and i get a(.050)=118(.030)(.050/.030) =118<<i don't...
18. ### Need a little help!

I haven't learned angular velocity yet so I'm not sure how to relate the two. I was trying to use the formula for centrepitual acceleration A=v^2/r I then set up the equation for the first distance from the center of the CD and then set up a second equation for the second distance from the...
19. ### Need a little help!

Va means velocity times the acceleration
20. ### Need a little help!

Homework Statement A computer is reading data from a rotating CD-ROM. At a point that is 0.030 m from the center of the disc, the centripetal acceleration is 118 m/s2. What is the centripetal acceleration at a point that is 0.050 m from the center of the disc? Homework Equations...

2.425 2.541
22. ### Two Dimensions

i haave exactly 4 mins to figure this out ive been working on this for hours plz help
23. ### Two Dimensions

i got you so now i need to find the magnitude and i use pythagorean theorem right?
24. ### Two Dimensions

i feel retarded that i don't understand this problem ive had the most problems with vectors than any other concept so far
25. ### Projectile Velocity

YES I GOT IT THAT TIME!! so i guess i have learned something today
26. ### Two Dimensions

o, okay so 2.7*cos43-x comp 2.7*sin43-y
27. ### Projectile Velocity

so if i calculated v_y right and t right my equation would be 68.727*.666+(-9.80m/s^2*(.666)^2)/(2)
28. ### Projectile Velocity

Well i think i messed up on the time it should be .666 i think b/c of the formula v=change of x/change in time?
29. ### Two Dimensions

isn't there only one x-component so it would =.45 and one y component=.70<<these answers seem too obvious am I missing something?
30. ### Projectile Velocity

V_y=75m/s*sin(66.4) then I use the formula (V_y*t-9.80*t^2)/2 then subtract that answer from 11m right?<<<am I'm on the right track I just keep getting a very large numbers for answers